A am to construct a polynomial with given information:
real roots $-1$ (With multiplicity 2 and 1) and $(2,f(2))=(2,4)$
The answer is: $f(x)=\frac{4}{9}(x^3+x^2-x-1)$
This is an extension exercise at the end of a book. In this particular case I don't even know how to get started or how to approach, I cannot see the bigger picture. For this reason I have not shown steps already taken or what I have tried already.
How can I arrive at $f(x)=\frac{4}{9}(x^3+x^2-x-1)$? Granular, baby steps greatly appreciated where possible.
Best Answer
Here is how to find a cubic polynomial. Let $$f(x)=A(x+1)^2(x+d)$$
where $d \ne 1$, $A \ne 0$.
When $x=2$, we have $4=9A(2+d)$
$$d=\frac{4}{9A}-2$$
$$f(x) = A(x+1)^2(x+\frac{4}{9A}-2)$$
Try to pick a suitable value of $A$ to get what you want.
Actually if we consider $f(x)=B(x+1)^2$, when $x=2$, we have $4=9B$ satisfy the condition unless I misinterpreted your question.
If the question says there is a root with multiplicity $2$ and another root with multiplicity $1$, then the polynomial is at least cubic.